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Babylonian method

Using the Babylonian method you can approximate the value of a positive square root x = √a by iterations of

 


Explanation

We want to calculate x = √a and search for an equivalent equation to do so

We can make an iteration of this, where a always lies between xn and a / xn. Then you can assume that the average of these two values gives a better value, so that

 


Example 1

If we start for x = √2 with a = 2 and x0 = 2, we get the following iterations

That's even more accurate than the value 1,414213562… your calculator shows.

 


Example 2

If we start for x = √9 with a = 9 and x0 = 2, we get the following iterations

If we start with x0 = 3, then the result is immediately

because 9 = 3, but we already knew that.

 


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